Equivalence of Definitions of Semiring of Sets/Definition 1 implies Definition 2

Theorem

Let $\SS$ be a system of sets satisfying the semiring of sets axioms:

\((1)\)	$:$							\(\ds \O \in \SS \)
\((2)\)	$:$		$\cap$-stable	\(\ds \forall A, B \in \SS:\)				\(\ds A \cap B \in \SS \)
\((3)\)	$:$			\(\ds \forall A, A_1 \in \SS : A_1 \subseteq A:\)				$\exists n \in \N$ and pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

Then $\SS$ satisfies the semiring of sets axioms:

\((1)\)	$:$							\(\ds \O \in \SS \)
\((2)\)	$:$		$\cap$-stable	\(\ds \forall A, B \in \SS:\)				\(\ds A \cap B \in \SS \)
\((3')\)	$:$			\(\ds \forall A, B \in \SS:\)				$\exists n \in \N$ and pairwise disjoint sets $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

Proof

Let $\SS$ be a system of sets satisfying the axioms:

\((1)\)	$:$							\(\ds \O \in \SS \)
\((2)\)	$:$		$\cap$-stable	\(\ds \forall A, B \in \SS:\)				\(\ds A \cap B \in \SS \)
\((3)\)	$:$			\(\ds \forall A, A_1 \in \SS : A_1 \subseteq A:\)				$\exists n \in \N$ and pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

It remains to be shown that $\SS$ satisfies the axiom

\((3')\)

$:$

\(\ds \forall A, B \in \SS:\)

$\exists n \in \N$ and pairwise disjoint sets $A_1, A_2, A_3, \ldots, A_n \in \SS : \ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$

Let $A, B \in \SS$.

Let $A_1 = A \cap B$.

By axiom $(2)$:

$A_1 \in \SS$

From Intersection is Subset:

$A_1 \subseteq A$

By axiom $(3)$:

$\exists$ a finite sequence of pairwise disjoint sets $A_2, A_3, \ldots, A_n \in \SS : \ds A = \bigcup_{k \mathop = 1}^n A_k$

Then:

\(\ds A \setminus B\)

\(=\)

\(\ds A \setminus \paren {A \cap B}\)

Set Difference with Intersection is Difference

\(\ds \)

\(=\)

\(\ds A \setminus A_1\)

\(\ds \)

\(=\)

\(\ds \paren {\bigcup_{k \mathop = 1}^n A_k} \setminus A_1\)

\(\ds \)

\(=\)

\(\ds \bigcup_{k \mathop = 1}^n \paren {A_k \setminus A_1}\)

Set Difference is Right Distributive over Union

\(\ds \)

\(=\)

\(\ds \bigcup_{k \mathop = 2}^n \paren {A_k \setminus A_1}\)

Set Difference with Self is Empty Set and Union with Empty Set

\(\ds \)

\(=\)

\(\ds \bigcup_{k \mathop = 2}^n A_k\)

Set Difference with Disjoint Set

As $A$ and $B$ were arbitrary, then $\SS$ satisfies axiom $(3')$

The result follows

$\blacksquare$